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I want to initialize a 5x5 array with 0 as content of each array index. But whenever i run my code, i get segmentation fault(core dumped error). Can you please help me what is the problem with my code. My code is as follows.

#include <stdio.h>
int main()
{
    int a[5][5];
    int i,j;
    for(i=0; i<=5; i++)
    {
        for(j=0; j<=5; j++)
        {
            a[i][j]=0;
        }
    }
}
Ayesha Javed
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    You have an array of 5 rows (indexed 0..4) and 5 cols (indexed 0..4), but you are accessing 6 rows (indexes 0..5) and 6 cols (indexes 0..5). By the way, you may initialize the array using `int a[5][5] = { 0 };`. – ikegami May 12 '20 at 16:52
  • An a[5] array is indexed from 0 to 4. Your loop stops at ```i<=5``` so you try to get ```a[5]``` -> Out of bounds. Stop the loop with ```i<5```. Idem for ```j``` – Olivier Depriester May 12 '20 at 16:53
  • You have out of bound access: the index range is from 0 to 4. So the fix the condition: `<=5` -> `<5`. Btw, in order to initialize the entire array to zero, you can instead do: `int a[5][5] = {0};`. – P.P May 12 '20 at 16:53

4 Answers4

3

You have an array of 5 rows (indexed 0..4) and 5 cols (indexed 0..4), but you are accessing 6 rows (indexes 0..5) and 6 cols (indexes 0..5). You need to adjust the bounds on your loop.

#define ROWS 5
#define COLS 5

int a[ROWS][COLS];
for (int i=0; i<ROWS; ++i) {
    for (int j=0; j<COLS; ++j) {
        a[i][j] = 0;
    }
}

That said, you could simply use the following to initialize the array:

#define ROWS 5
#define COLS 5

int a[ROWS][COLS] = { 0 };

Note that I used names rather than hardcoding the numbers everywhere. This is far more readable and far less error-prone.

ikegami
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2

For arrays in C, indexing starts from 0.

When we declare an array a[5], which obviously indicates that we are allocating 5 memory space of some data types for a, one way to access those memory address is to use the indexes. As I have stated earlier that in C, indexing starts from 0, those 5 space would be:

  1     2     3     4     5    <--------- Actual position
a[0]  a[1]  a[2]  a[3]  a[4]   <--------- Array index

So, when we will try to access a[5] for above example, that will lead to segmentation fault.

Similarly, for your case, you should correct your for-loop like below:

for (i = 0; i < 5; i++) {
    for (j = 0; j < 5; j++) {
        a[i][j] = 0;
    }
}
UkFLSUI
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0

The problem is <= insted of <

It will do the loop 6 times ,and set 0 in each a[i][j] in the 2d array

It will start from a[0][0] to a[5][5]

but u gave hem a memory of 5 cubes(0,1,2,3,4 its start from 0 , not 1) , and the loop set 6 cubes(0,1,2,3,4,5) but 5 is not in the array , so u have a segmentation fault.

Ardent Coder
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Eli Revach
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-1

Your mistake:

for(i=0; i<=5; i++) // accessed extra element where there are only 5 (i < 5)
    for(j=0; j<=5; j++) // same with this, correct: (j < 5)
        a[i][j]=0;

Correct solution is:

for(i = 0; i < 5; i++)
        for(j = 0; j < 5; j++)
            a[i][j]=0;

Use:

int a[5][5] = {0};

to initialize multidimensional arrays to zero.

More concepts could be found in this thread.

Rohan Bari
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